Question

Let $a,b$ and $c$ be the length of sides of a triangle $ABC$ such that $\frac{a+b}{7}=\frac{b+c}{8}=\frac{c+a}{9}$. If $r$ and $R$ are the radius of incircle and radius of circumcircle of the triangle $ABC$, respectively, then the value of $\frac{R}{r}$ is equal to

• A. $\frac{5}{2}$
• B. 2
• C. $\frac{3}{2}$
• D. 1

Answer 1

Answer: (A)

$\frac{a+b}{7}=\frac{b+c}{8}=\frac{c+a}{9}=\lambda$
$a+b=7\lambda ,b+c=8\lambda ,a+c=9\lambda$
$\Rightarrow a+b+c=12\lambda$
Now $a=4\lambda ,b=3\lambda ,c=5\lambda$
$\because c^2=b^2+a^2$
$\angle C=90^{\circ }$
$\Delta =\frac{1}{2}ab\sin C=\frac{1}{2}ab$
$\frac{R}{r}=\frac{c}{2\sin C}\times \frac{s}{\Delta }=\frac{c}{2}\times \frac{6\lambda }{\frac{1}{2}ab}=\frac{c}{ab}\times 6\lambda =\frac{5}{2}$